Problem: Is ${188713}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {188713}= &&{1}\cdot100000+ \\&&{8}\cdot10000+ \\&&{8}\cdot1000+ \\&&{7}\cdot100+ \\&&{1}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {188713}= &&{1}(99999+1)+ \\&&{8}(9999+1)+ \\&&{8}(999+1)+ \\&&{7}(99+1)+ \\&&{1}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {188713}= &&\gray{1\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {1}+{8}+{8}+{7}+{1}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${188713}$ is divisible by $3$ if ${ 1}+{8}+{8}+{7}+{1}+{3}$ is divisible by $3$ Add the digits of ${188713}$ $ {1}+{8}+{8}+{7}+{1}+{3} = {28} $ If ${28}$ is divisible by $3$ , then ${188713}$ must also be divisible by $3$ ${28}$ is not divisible by $3$, therefore ${188713}$ must not be divisible by $3$.